SOLUTION: The length of a rectangle is 4 units less than twice the width. If the length is decreased by 3 units and the width is increased by 1 unit, the area is decreased by 16 square unit
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Question 289656: The length of a rectangle is 4 units less than twice the width. If the length is decreased by 3 units and the width is increased by 1 unit, the area is decreased by 16 square units. If w is the original width, which equation must be true?
A. (2w-3)(w-3)=w(2w-4)-16
B. 2(2w-3)+2(w-3)=2w+2(2w-4)-16
C. (2w-7)(w+1)=w(2w-4)-16
D. 2(2w-7)+2(w+1)=2w+(2w-4)-16 Answer by richwmiller(17219) (Show Source):
You can put this solution on YOUR website! l=-4+2w
l=2w-4
new l ((2w-4)-3)
new (w+1)
((2w-4)-3)*(w+1)=(2w-4)*(w)-16
(2w-7)*(w+1)=(2w-4)*(w)-16
C. answer
